1. ## differential analysis

Find an example of a differentiable function
f : R R such that f'(0) = 1 > 0 but f is not monotonic increasing on any interval (0,delta).
Suppose that
f : R R has derivatives of all orders. Prove that the same

is true of
F(x) := exp (f(x)).

not sure on either of these two questions..

2. Originally Posted by James0502
Find an example of a differentiable function
f : R R such that f'(0) = 1 > 0 but f is not monotonic increasing on any interval (0,delta).
Suppose that
f : R R has derivatives of all orders. Prove that the same

is true of
F(x) := exp (f(x)).

not sure on either of these two questions..

For the first one, the following link might be helpful Weierstrass function - Wikipedia, the free encyclopedia, and for the second one, you may refer to http://www.mathhelpforum.com/math-he...-new-post.html
About the first one I really wonder if there is an explicit answer, as I remember such a discussion was made previously .

3. Originally Posted by James0502
Find an example of a differentiable function
f : R R such that f'(0) = 1 > 0 but f is not monotonic increasing on any interval (0,delta).
Suppose that
f : R R has derivatives of all orders. Prove that the same

is true of
F(x) := exp (f(x)).

not sure on either of these two questions..

For the first question, you need $\displaystyle f$ to be differentiable everywhere, but $\displaystyle f'$ must be discontinuous at 0 (otherwise, it would be positive near 0 and $\displaystyle f$ would be increasing).
A quite famous example of a differentiable function which is not continuously differentiable at 0 is the following: $\displaystyle g(x)=x^2\sin\frac{1}{x}$ with $\displaystyle g(0)=0$. However, $\displaystyle g'(0)=0$, so we should change it a little... You can prove that the function $\displaystyle f:x\mapsto f(x)=x + 2x^2\sin\frac{1}{x}$ with $\displaystyle f(0)=0$ is an answer to your problem. The "2" is probably optional but may simplify the proof.
You have to prove that $\displaystyle f$ is differentiable on $\displaystyle (0,+\infty)$ (straightforward), continuous at 0, differentiable at 0 (look at $\displaystyle \frac{f(x)}{x}$), that $\displaystyle f'(0)=1$ and that $\displaystyle f'(x)$ has not a constant sign on any neighbourhood $\displaystyle (0,h)$ of 0.